How do you factor #x^3+2x^2-8x-16?

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Answer 1 · 2015-05-09T20:11:57

Try x = -2.
$f (- 2) = - 8 + 8 + 16 - 16 = 0 .$ $f(-2)= -8 + 8 +16 - 16 = 0.$ Then f(x) can be divided by (x + 2)

$f (x) = (x + 2) (x^{2} - 8) .$ $f(x) = (x + 2)(x^2 - 8).$

Answer 2 · 2015-05-09T20:26:03

The answer is $(x^{2} - 8) (x + 4)$ $(x^2-8)(x+4)$ .

Factor $x^{3} + 2 x^{2} - 8 x - 16$ $x^3+2x^2-8x-16$ .

Since there is no common factor for the polynomial, factor by grouping.

$x^{3} + 2 x^{2} - (8 x - 16)$ $color(red)(x^3+2x^2)color(blue)(-(8x-16)$

Factor $x^{2}$ $color(red)x^2$ out of the first term.

$x^{2} (x + 2)$ $color(red)(x^2(x+2)$

Factor $- 8$ $color(blue)(-8$ out of the second term.

$- 8 (x + 2)$ $color(blue)(-8(x+2)$

Put the two sets of terms back together.

$x^{2} (x + 2) - 8 (x + 2)$ $color(red)(x^2(x+2)color(blue)(-8(x+2)$

$(x + 2)$ $color(purple)((x+2)$ is now the common factor.

Factor out $(x + 2)$ $color(purple)((x+2)$ .

Final answer = $(x^{2} - 8) (x + 2)$ $color(purple)((x^2-8)(x+2)$